Hydrothermal Geyser Formation (2D): Boiling Point at Depth, Heating Cycle and Eruption Height

Presets

Live readouts

Graphs

Chamber water temperature and boiling point at depth

Jet height of each eruption

Eruption cycle length against heat supply

Numerical checks

running...

How to use

The orange curve is the chamber water temperature, the red dashed line the boiling point at the chamber depth: when they meet, the geyser erupts. Try: (1) Hot recharge: warmer inflow shortens the wait; (2) Deep and slow: a higher boiling point means more superheat, a taller jet and a longer interval; (3) Dormant spring: the heat loss through the rock balances the supply and the water never reaches boiling; (4) in the right-hand graph, the interval falls as the heat supply rises; (5) raise the speed to watch many eruptions. Space pauses, R resets.

The science

What this lab shows. A geyser is a hot-water system in which boiling is delayed by the weight of the water above. Hot rock heats a water-filled chamber at the bottom of a narrow conduit. Pressure at that depth raises the boiling point; when the water finally reaches it, a little of the column flashes to steam, expels the water above, the pressure falls, and the rest of the superheated water flashes in a violent eruption. Cool groundwater then refills the chamber and the cycle restarts.

1. Boiling point rises with depth. The pressure at depth H in a water column is p = p_atm + ρ g H, and the saturation temperature follows the Antoine relation (fit to steam tables)

ln p_mmHg = 18.3036 - 3816.44 / (T + 227.02)

so water at 22 m depth (about 3.2 bar) boils only at about 135 C, not 100 C. That extra 35 K is the energy store that drives the eruption.

2. Heating the chamber. With heat input Q, chamber water mass m, heat capacity C = m c_w and a loss coefficient UA to the cooler surroundings:

C dT/dt = Q - UA (T - T_env) ⇒ T(t) = T_∞ - (T_∞ - T₀) e^{-t UA / C}, T_∞ = T_env + Q/UA

The waiting time is the time for T to reach the boiling point: t = -(C/UA) ln[(T_∞ - T_b)/(T_∞ - T₀)]. If T_∞ is below T_b the system never erupts (a quiet hot spring).

3. The eruption. When the pressure drops to 1 atm the superheated water flashes; the steam mass fraction is x = c_w (T_b - 100 C) / h_fg (h_fg = 2257 kJ/kg). A small fraction η of that flash energy becomes kinetic energy of the jet: ½ v² = η x h_fg, so the jet height is v²/(2g). The fraction φ of the chamber is ejected; the chamber is then refilled with recharge water at T_in and the remaining hot water, giving the new starting temperature T₀ = (1 - φ) 100 + φ T_in.

Checks (run in the page). (1) and (2) the boiling point at 1 atm and at 10 bar against steam tables; (3) numerical heating against the closed-form exponential; (4) the numerical time to boiling against the analytic formula; (5) the heat balance Q t = C ΔT + the integral of the loss; (6) the jet height v²/(2g) against a numerically integrated ballistic rise.

Simplifications. Real geysers have complicated, irregular plumbing: bubble traps, side chambers, two-phase flow in the conduit and recharge that depends on rainfall and neighbouring springs, so real intervals are irregular. Here the chamber is one well-mixed lump; the fraction φ and the jet efficiency η are adjustable constants (the value 0.002 is chosen so that the default produces an Old Faithful-sized jet, it is not derived); dissolved gases, silica deposition and the surface pool are not modelled. The model illustrates the mechanism and the scaling, not the behaviour of any particular geyser.